# Evaluating $\int\frac{dx}{\sqrt{4-9x^2}}$ with different trig substitutions ($\sin$ vs $\cos$) gives different results

I am trying to solve the following integral with trig substitutions. However, I get a different answer for two substitutions that should yield the same result. $$\int\frac{dx}{\sqrt{4-9x^2}}$$

• For the first trig sub, I set $$9x^2 = 4\cos^2\theta$$. This simplifies to: $$x = \frac{2}{3}\cos\theta$$, and $$dx = -\frac{2}{3}\sin\theta$$. Substituting in, I get: $$\int\frac{-2\sin\theta}{6\sin\theta} = -\frac{\theta}{3} = -\frac{1}{3}\,\cos^{-1}\left(\frac{3x}{2}\right)+C \tag{1}$$

• For the second trig sub, I set $$9x^2 = 4\sin^2\theta$$. This simplifies to: $$x = \frac{2}{3}\sin\theta$$, and $$dx = \frac{2}{3}\cos\theta$$. Substituting in, I get: $$\int\frac{2\cos\theta}{6\cos\theta} = \frac{\theta}{3} = \frac{1}{3}\,\sin^{-1}\left(\frac{3x}{2}\right)+C \tag{2}$$

My question is:

Why do these two trig substitutions yield different results graphically? Shouldn't they result in the same graph?

Since $$(-\arcsin)'(x)=\arccos'(x)=-\dfrac1{\sqrt{1-x^2}}$$, you got twice the same thing.
• Because their difference is a constant:$$\arccos(x)+\arcsin(x)=\frac\pi2.$$So, your $C$'s are actually two distinct constants. Commented Mar 24, 2019 at 16:03